# Hermitian matrices in an arbitrary inner product

I was wondering if anyone can give me any hints for how to prove the following proposition:

Proposition: Let $V = F^n$ and let $A\in M_{n\times n}(F)$. Prove that $\langle A^* x,y \rangle = \langle x, A y \rangle$.

The problem that I am having is how to prove this for an arbitry inner product on $F^n$, an arbitrary field. Some textbooks take this as an axiom or the definition of the adjoint of $A$, but the book that I got this from does not (Linear Algebra by Friedberg).

• What is the definition of $A^*$? What is $F$? – user251257 Apr 29 '17 at 3:35
• $A^*$ is the adjoint of the matrix, and $F$ is the field over which the inner product is defined. This could be the real numbers, complex field, etc. – OscarAraiza Apr 29 '17 at 3:38
• any inner product of $V$ can be expressed by a positive definite hermitian /symmetric matrix $B$ as in $\langle x, y \rangle = x^* B y$. So i doubt it will work if $A$ and $B$ don't commute. – user251257 Apr 29 '17 at 3:44
• Which edition of the book are you using, and what page is this on? I suspect the reference is to the standard inner product on $F^n$, namely $\langle x,y \rangle = y^{*} x$. – user49640 Apr 29 '17 at 5:15

This is true only for $V$ with the standard inner product.

Counter-example:

Assume $V=\mathbb{R}^2$ and we define the inner product in the following way: $$\langle x | y\rangle = x^T \begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} y$$

Lets take for simplicity $x = \begin{pmatrix} 1\\0 \end{pmatrix} , y = \begin{pmatrix} 0\\1 \end{pmatrix}$

Lets take $A$ to be $\begin{pmatrix}1 & 2\\3 & 4 \end{pmatrix}$

Now some calculations: $$\langle x | Ay \rangle = x^T\begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} \begin{pmatrix}1 & 2\\3 & 4 \end{pmatrix} y = x^T\begin{pmatrix}1+\frac{3\sqrt{3}}{2} & 2(1+\sqrt{3})\\\frac{1}{2}(6+\sqrt{3}) & 4+\sqrt{3} \end{pmatrix} y = 2(1+\sqrt{3})$$ $$\langle A^*x | y \rangle = \left (\begin{pmatrix}1 & 3\\2 & 4 \end{pmatrix} x\right )^T\begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} y = \begin{pmatrix} 1& 2 \end{pmatrix}\begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} \begin{pmatrix} 0\\ 1 \end{pmatrix} = 2 + \frac{\sqrt{3}}{2}$$

$$\langle x | Ay \rangle \neq \langle A^*x | y \rangle$$

• Nice observation. – srijan Apr 29 '17 at 10:08
• I wouldn't go so far to claim, that it is only for the standard inner product. Every positive multiple of it will also do. – user251257 Apr 29 '17 at 10:28
• @user251257 you're right. – Ofek Gillon Apr 29 '17 at 10:45

As pointed out in the comments and the answer written above, following answer may give you a little more detail. I have taken the real inner product space.

Consider the inner product space $V = \mathbb{R}^n$ with the standard inner product defined as follows:

Let $x = (x_1, x_2, x_3, \ldots x_n)$, and $y = (y_1, y_2, y_3, \ldots y_n)$ are any two arbitrary vectors in $\mathbb{R}^n$. We define the standard inner product in $\mathbb{R}^n$, denoted by $\langle x, y\rangle$ by

$\langle x, y\rangle = x_1 y_1 + x_2 y_2 + \ldots + x_n y_n$.

How to define the dot product between $x$ and $y$ if we consider them as a column vectors of $\mathbb{R}^n$ having size $n\times 1$?. It can be defined as

$x.y = x^t. y = x_1 y_1 + x_2 y_2 + \ldots + x_n y_n$, which is same as the standard inner product definition in $\mathbb{R}^n$. Thus we can write

$\langle x, y\rangle = x.y = x^t. y$

LHS = $\langle A^t x,y \rangle = (A^t x)^t y = x^t Ay$
RHS = $\langle x, A y \rangle = x^t A y$